MCQ
If $\text{z}=\frac{1}{1-\cos\theta-\text{i}\sin\theta},$ then $\text{Re(z)}=$
  • A
    $0$
  • $\frac{1}{2}$
  • C
    $\cot\frac{\theta}{2}$
  • D
    $\frac{1}{2}\cot\frac{\theta}{2}$

Answer

Correct option: B.
$\frac{1}{2}$
$\text{z}=\frac{1}{1-\cos\theta-\text{i}\sin\theta}$
$\text{z}=\frac{1}{1-\cos\theta-\text{i}\sin\theta}\times\frac{1-\cos\theta+\text{i}\sin\theta}{1-\cos\theta+\text{i}\sin\theta}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{(1-\cos\theta)^2-(\text{i}\sin\theta)^2}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{1+\cos^2\theta-2\cos\theta+\text{i}\sin^2\theta}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{1+1-2\cos\theta}$
$\Rightarrow\text{z}=\frac{1-\cos\theta+\text{i}\sin\theta}{2(1-\cos\theta)}$
$\Rightarrow\text{Re(z)}=\frac{(1-\cos\theta)}{2(1-\cos\theta)}=\frac{1}{2}$

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